similar to: pseudo-R by hand

Hi all; I recently have been used 'maxLik' function for maximizing G2StNV178 function with gradient function gradlik; for receiving this goal, I write the following program; but I have been seen an error  in calling gradient  function; The maxLik function can't enter gradlik function (definition of gradient function); I guess my mistake is in line ******** ,that the vector  ‘h’ is

Dear R users, i would like to transform the following function from R-code to C-code and call it from R in order to speed up the computation because in my other functions this function is called many times. `dgcpois` <- function(z, lambda1, lambda2) { `f1` <- function(alpha, lambda1, lambda2) return(exp(log(lambda1) * (alpha - 1) - lambda2 * lgamma(alpha))) `f2` <-

How to convert the following ms() in Splus to Optim in R? The "Calc" function is also attached. ms(~ Calc(a.init, B, v, off, d, P.a, lambda.a, P.y, lambda.y, 10^(-8), FALSE, 20, TRUE)$Bic, start = list(lambda.a = 0.5, lambda.y = 240), control = list(maxiter = 10, tol = 0.1)) Calc <- function(A.INIT., X., V., OFF., D., P1., LAMBDA1., P2., LAMBDA2., TOL., MONITOR.,

Dear R-users, I am trying to translate a matlab code for calculating the Local Whittle estimator in time series with long memory originally written by Shimotsu and available free in his webpage ( http://www.econ.queensu.ca/pub/faculty/shimotsu/ ) The Matlab code is ======================================================================================= function[r] = whittle(d,x,m) % WHITTLE.M

I use "while" loop but it produces an errro. I have no idea about this. Error in "[<-"(`*tmp*`, i, value = numeric(0)) : nothing to replace with The problem description is The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0

I have an error for a simple optimization problem. Is there anyone knowing about this error? lambda1=-9 lambda2=-6 L<-function(a){ s2i2f<-(exp(-lambda1*(250^a)-lambda2*(275^a-250^a)) -exp(-lambda1*(250^a)-lambda2*(300^a-250^a))) logl<-log(s2i2f) return(-logl)} optim(1,L) Error in optim(1, L) : function cannot be evaluated at initial parameters Thank you in advance -- View this

The following R code using different packages gives the same results for a simple logistic regression without regularization, but different results with regularization. This may just be a matter of different scaling of the regularization parameters, but if anyone familiar with these packages has insight into why the results differ, I'd appreciate hearing about it. I'm new to

Rainer M Krug writes: > Thorsten Wiegand used in his paper Wiegand T., and K. A. Moloney 2004. > Rings, circles and null-models for point pattern analysis in ecology. > Oikos 104: 209-229 a statistic he called O-Ring statistic which is > similar to Ripley's K, only that it uses rings instead of circles. > > http://www.oesa.ufz.de/towi/towi_programita.html#ring

Dear R and S users, I've been working on fitting finite mixture of negative exponential distributions using maximum likelihood based on the example given in MASS. So far I had much success in fitting two components. The problem started when I tried to extend the procedure to fit three components. More specifically, likelihood = sum( ln(c1*exp(-x/lambda1)/lambda1 + c2*exp(-x/lambda2)/lambda2

Dear all I used R2WinBUGS package's bugs() function to generate MCMC results. Then I tried to save the simulation draws in R, using read.bugs() function. Here is a simple test: ###################### library(coda) library(R2WinBUGS) #fake some data to test beta0=1 beta1=1.5 beta2=-1 beta3=2 N=200 x1=rnorm(N, mean=0,sd=1) x2=rnorm(N, mean=0,sd=1) x3=rnorm(N, mean=0,sd=1) lambda2= exp(beta0+

On Tuesday 16 December 2008 01:03:36 Evan Cheng wrote: > FYI. http://leonardo-m.livejournal.com/73732.html > > If anyone is motivated, please file bugs for the losing cases. Also, > it might make sense to incorporate the tests into our nightly tester > test suite. FWIW, I just ported my ray tracer benchmark to C and found that llvm-gcc gives much worse performance than gcc on x86

Dear all, can somebody please help me how to calculate "McFadden R-square" for a LOGIT model? Corresponding definition can be found here: http://publib.boulder.ibm.com/infocenter/spssstat/v20r0m0/index.jsp?topic=%2Fcom.ibm.spss.statistics.help%2Falg_plum_statistics_rsq_mcfadden.htm Here is my data: Data <- structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1,

Hi All, I am running a generalized linear model with gamma distribution in R (glm, family=gamma ) for my data (gene expression as response variable and few predictors). I want to calculate r-squared for this model. I have been reading online about it and found there are multiple formulas for calculating R2 (psuedo) for glm (in R) with gaussian (r2 from linear model), logistic regression

Dear all, I have two functions (f1, f2) and 4 unknown parameters (p1, p2, p3, p4). Both f1 and f2 are functions of p1, p2, and p3, denoted by f1(p1, p2, p3) and f2(p1,p2,p3) respectively. The goal is to maximize f1(p1, p2, p3) subject to two constraints: (1) c = k1*p4/(k1*p4+(1-k1)*f1(p1,p2,p3)), where c and k1 are some known constants (2) p4 = f2(p1, p2, p3) In addition, each parameter

Dear R users, I am using "enet" package in R for applying "elastic net" method. In elastic net, two penalities are applied one is lambda1 for LASSO and lambda2 for ridge ( zou, 2005) penalty. But while running the analysis, I realised tht, I optimised only one lambda. ( even when I looked at the example in R, they used only one penality) So, I am

Hello, The likelihood includes two parameters to be estimated: lambda (=beta0+beta1*x) and alpha. The algorithm for the estimation is as following: 1) with alpha=0, estimate lambda (estimate beta0 and beta1 via GLM) 2) with lambda, estimate alpha via ML estimation 3) with updataed alpha, replicate 1) and 2) until alpha is converged to a value I coded 1) and 2) (it works), but faced some

Hello. I am running 9 poisson regressions with 5 predictors each, using glm with family=gaussian. Gaussian distribution fits better than linear regression on fit indices, and also for theoretical reasons (e.g. the dependent variables are counts, and the distribution is highly positively skewed). I want to determine pseudo R^2 now. However, using the pR2() of the pscl package offers drastically

Hi, I have recently been attempting to find the LD50 from two predicted fits (For male and females) in a Generalised linear model which models the effect of both sex + logdose (and sex*logdose interaction) on proportion survival (formula = y ~ ldose * sex, family = "binomial", data = dat (y is the survival data)). I can obtain the LD50 for females using the dose.p() command in the MASS

I am interested Andrea is whether you ever established why your R2 was 1. I have had a similar situation previously. My main issue though, which I'd be v grateful for advice on, is why I am obtaining such negative values -0.3 for Somers Dxy using validate.cph from the Design package given my value of Nagelkerke R2 is not so low 13.2%. I have this output when fitting 6 variables all with

Running a binary logit model on the data df <- data.frame(y=sample(letters[1:3], 100, repl=T), x=rnorm(100)) reveals some residual deviance: summary(glm(y ~ ., data=df, family=binomial("logit"))) However, running a multinomial model on that data (multinom, nnet) reveals a residual deviance: summary(multinom(y ~ ., data=df)) On page 203, the MASS book says that "here the